Filling a byte array in Java

Question

Hey all!

For part of a project I'm working on I am implementing a RTPpacket where I have to fill the header array of byte with RTP header fields.

  //size of the RTP header:
  static int HEADER_SIZE = 12; // bytes

  //Fields that compose the RTP header
  public int Version; // 2 bits
  public int Padding; // 1 bit
  public int Extension; // 1 bit
  public int CC; // 4 bits
  public int Marker; // 1 bit
  public int PayloadType; // 7 bits
  public int SequenceNumber; // 16 bits
  public int TimeStamp; // 32 bits
  public int Ssrc; // 32 bits

  //Bitstream of the RTP header
  public byte[] header = new byte[ HEADER_SIZE ];

This was my approach:

/*      
 * bits 0-1: Version
 * bit    2: Padding 
 * bit    3: Extension
 * bits 4-7: CC
 */
header[0] = new Integer( (Version << 6)|(Padding << 5)|(Extension << 6)|CC ).byteValue();

/* 
 * bit    0: Marker
 * bits 1-7: PayloadType
 */
header[1] = new Integer( (Marker << 7)|PayloadType ).byteValue();

/* SequenceNumber takes 2 bytes = 16 bits */
header[2] = new Integer( SequenceNumber >> 8 ).byteValue();
header[3] = new Integer( SequenceNumber ).byteValue();

/* TimeStamp takes 4 bytes = 32 bits */
for ( int i = 0; i < 4; i++ )
    header[7-i] = new Integer( TimeStamp >> (8*i) ).byteValue();

/* Ssrc takes 4 bytes = 32 bits */
for ( int i = 0; i < 4; i++ )
    header[11-i] = new Integer( Ssrc >> (8*i) ).byteValue();

Any other, maybe 'better' ways to do this?

Answer 1

You can convert an int directly to a byte in Java, without having to create an Integer object. An explicit cast is required, because a byte has a narrower range of possible values than an int. For instance:

header[1] = (byte) (Marker << 7 | PayloadType);

Answer 2

There is one problem with such data. Usually protocols use unsigned bytes there and Java has signed bytes. So, for correct byte array fill I usually use such construct:

bytearray[index] = (byte) ((some integer-result calculation) & 0xff);

Simple casting to byte type won't work correctly.

Update. "& 0xff" is not needed here. Simple cast will work.

Answer 3

I think I would use a ByteBuffer

ByteBuffer buf = ByteBuffer.wrap(header);
buf.setOrder(ByteOrder.BIG_ENDIAN);
buf.put((byte)((Version << 6)|(Padding << 5)|(Extension << 6)|CC));
buf.put((byte)((Marker << 7)|PayloadType));
buf.put((short)SequenceNumber);
buf.put(TimeStamp);
buf.put(Ssrc);

Answer 4

in addition to the presented answers give a try to Preon

Answer 5

With Preon, the RtpHeader could be represented like this:

public class RtpHeader {

    @BoundNumber(size = "2")
    public int version;

    @Bound
    public boolean padding;

    @Bound
    public boolean extension;

    @BoundNumber(size="4")
    public int csrcCount;

    @Bound
    public boolean marker;

    @BoundNumber(size="7")
    public int payloadType;

    @BoundNumber(size="16", byteOrder = ByteOrder.BigEndian)
    public int sequenceNumber;

    @BoundNumber(size="32", byteOrder = ByteOrder.BigEndian)
    public int timestamp;

    @BoundNumber(size="32", byteOrder = ByteOrder.BigEndian)
    public int synchronizationSource;

    @BoundList(size="csrcCount")
    public int[] csrcs; 

}

Encoding this to bytes could be as simple as this:

    Codec<RtpHeader> codec = Codecs.create(RtpHeader.class);
    RtpHeader header = new RtpHeader();
    ... // Setting header values
    OutputStream out = ...;
    Codecs.encode(header, codec, out);

However, bear in mind that encoding in Preon is still in early stages. It seems to work, for this particular case, but I am not going to make any guarantees.

The benefit of using Preon is obviously the fact that you don't have to bother writing all of the encoding and decoding logic yourself.